Can the Auger effect cause a second electron to be just excited instead of ionised and emitted from the atom? From what I understand, the Auger effect is usually defined as when an electron deexcites but instead of releasing its change in binding energy as a photon, it transfers it as kinetic energy to another electron which, if greater than its binding energy, will cause this second electron to be emitted from the atom.
My question is why is this process defined with the second electron being emitted from the atom instead of just excited to a higher energy state sometimes.
My guess is maybe it has something to do with entropy and the fact that there are so many more possible states for the second electron final state if it is emitted that maybe only in this case will this process actually occur (instead of just emitting a photon as usual).
 A: Let me first note that Auger process is due to Coulomb interaction between electrons, so it may be beneficial to think of it in terms of the Fermi golden rule:
$$
w_{i_1 i_2\rightarrow f_1 f_2}=\frac{2\pi}{\hbar}|\langle i_1, i_2 | V|f_1, f_2\rangle|^2\delta(\epsilon_{i_1} + \epsilon_{i_2} - \epsilon_{f_1} - \epsilon_{f_2})
$$.
One needs to be careful when calculating the matrix element to account properly for the exchange term, but what interests us at the moment is the energy conservation, codified here in the delta-function. If the first electron changes its energy between two levels:
$\epsilon_{i_1}-\epsilon_{f_1} = E_n-E_m$ than the second electron must increase its energy by the same amount: $\epsilon_{f_2}-\epsilon_{i_2} = E_n-E_m$.

*

*If, e.g., we were discussing Auger recombination in a semiconductor crystal, then satisfying this condition would be rather easy, since there are many states available in the condiction band. In an atom with discrete levels this however could be tricky, so being ejected into the continuum of states is the only option.

*Another consideration is the pure size of the energy change. If we take hydrogen-like spectrum
$$
E_n=-\frac{E_1}{n^2}.
$$
then
$$
E_n - E_1 = E_1\left(1-\frac{1}{n^2}\right) > \frac{1}{n^2} \text{ for any } n>1.
$$
A: 
My question is why is this process defined with the second electron being emitted from the atom instead of just excited to a higher energy state sometimes.

The atom is a unit tied up quantum mechanically . To observe transformations of an atom, there must be an interaction that can be measured. An emitted photon can be measured. An emitted electron can also be measured. If the whole process happens within an atom, there is no measurable/observable effect. An electron just going to a higher energy level can emit a photon when de-exciting , but there is no measurable way to determine that it comes from a transfer from a different electron, the way there is for an ejected electron to be identified with a different energy level:

Upon ejection, the kinetic energy of the Auger electron corresponds to the difference between the energy of the initial electronic transition into the vacancy and the ionization energy for the electron shell from which the Auger electron was ejected.

